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\begin{document}

In a general decimation procedure for an Ising-like Hamiltonian we may separate the spins into those to be retained $\odot$ and those to be ``decimated'' $\times$, we then have (schematically)

$$Z_N = \sum_\odot \sum_\times e^{-\beta \ham(\odot,\times)} = \sum_\odot e^{-\beta \ham'(\odot)}$$

where we may have to modify the original $\ham$ by the addition of a (physically unimportant) constant in order to write

$$e^{- \beta \ham'(\odot)} = \sum_\times e^{-\beta \ham(\odot,\times)}$$

which defines the ``renormalised Hamiltonian'' $\ham'(\odot)$ from which the $\times$ dependence has been eliminated. The coupling constants (say $K_i$) within $\ham$ will be changed by renormalisation to a new set $K_i'$ appearing within $\ham'$, and by performing the $\times$ sum we obtain the RG equation which relates $K_i$ to $K_i'$.

\begin{itemize}
\item

We begin the decimation procedure by modifying the original hamiltonian

$$ - \beta \ham = \sum_{i=1}^N K \sigma_i \sigma_{i+1}, \, K \geq 0$$

by the addition of a trivial field $K_0$

$$ - \beta \ham = \sum_{i=1}^N K \sigma_i \sigma_{i+1} + \sum_{i=1}^N K_0$$

and since only \emph{differences} in energy are physically significant, all observables are unchanged by this transformation.

Now the partition function is given by

$$Z_N = \sum_{\sigma_1 = \pm 1}\sum_{\sigma_2 = \pm 1}\cdots\sum_{\sigma_N = \pm 1} \exp \left\{ \sum_{i=1}^N K \sigma_i \sigma_{i+1} + \sum_{i=1}^N K_0 \right\} = \sum_{\{\sigma\}} \prod_{i=1}^N e^{w(\sigma_i,\sigma_{i+1})}$$

Under a $\lambda = 3$ decimation only every third spin is retained, and the intermediate spins summed out

$$Z_N = \sum_{\sigma_1 = \pm 1} \sum_{\sigma_4 = \pm 1} \cdots \sum_{\sigma_N = \pm 1} \left[ \sum_{\sigma_2 = \pm 1} \sum_{\sigma_3 = \pm 1} e^{w(\sigma_1,\sigma_2)} e^{w(\sigma_2,\sigma_3)} e^{w(\sigma_3,\sigma_4)} \right] \cdots$$
$$\cdots \left[ \sum_{\sigma_5 = \pm 1} \sum_{\sigma_6 = \pm 1} e^{w(\sigma_4,\sigma_5)} e^{w(\sigma_5,\sigma_6)} e^{w(\sigma_6,\sigma_7)} \right] \cdots $$
$$= \sum_{\sigma_1 = \pm 1} \sum_{\sigma_4 = \pm 1} \cdots \sum_{\sigma_N = \pm 1} e^{w'(\sigma_1,\sigma_4)} e^{w'(\sigma_4,\sigma_7)} \cdots$$

so the partition function sum retains the same form, but for a renormalised interaction $w'$. Since this is the same for every unit of 4 spins, we need only study one representative unit

$$e^{w'(\sigma_i,\sigma_{i+3})} = \sum_{\sigma_{i+1} = \pm 1} \sum_{\sigma_{i+2} = \pm 1} e^{w(\sigma_i,\sigma_{i+1})} e^{w(\sigma_{i+1},\sigma_{i+2})} e^{w(\sigma_{i+2},\sigma_{i+3})}$$

Now by our choice of $\ham$ we have

$$w(\sigma_i,\sigma_{i+1}) = K \sigma_i \sigma_{i+1} + K_0$$

and for the renormalised couplings we define

$$w'(\sigma_i,\sigma_{i+1}) = K' \sigma_i \sigma_{i+1} + K'_0$$

so we can write

$$e^{K'_0} e^{K' \sigma_i \sigma_{i+3}} = e^{3K_0}\sum_{\sigma_{i+1} = \pm 1} \sum_{\sigma_{i+2} = \pm 1} e^{K \sigma_i \sigma_{i+1}} e^{K \sigma_{i+1} \sigma_{i+2}} e^{K \sigma_{i+2} \sigma_{i+3}}$$

recalling the identity $e^{K \sigma \sigma'} = \cosh K (1 + \sigma \sigma' \tanh K)$ this becomes

$$e^{K'_0} e^{K' \sigma_i \sigma_{i+3}} = 2^2 e^{3K_0} \cosh^3 K \left( 1 + \sigma_i \sigma_{i+3} \tanh^3 K \right)$$

applying the same identity to the LHS and equating with the above

$$e^{K'_0} \cosh K' (1 + \sigma_i \sigma_{i+3} \tanh K') = 2^2 e^{3K_0} \cosh^3 K \left( 1 + \sigma_i \sigma_{i+3} \tanh^3 K \right)$$

which is valid for arbitrary $\sigma_i \sigma_{i+3} = \pm 1$ so we can write 2 equations

$$e^{K'_0} \cosh K' ( 1 + \tanh K' ) = 2^2 e^{3K_0} \cosh^3 K ( 1 + \tanh^3 K )$$
$$e^{K'_0} \cosh K' ( 1 - \tanh K' ) = 2^2 e^{3K_0} \cosh^3 K ( 1 - \tanh^3 K )$$

the quotient of which gives

$$\frac{1 + \tanh K'}{1 - \tanh K'} = \frac{1 + \tanh^3 K}{1 - \tanh^3 K}$$

which is the RG equation for the ferromagnetic coupling $K$

$$K' = \tanh^{-1} ( \tanh^3 K )$$

and after some algebra we obtain for the trivial field

$$K'_0 = \log \left\{ \frac{2^2 e^{3K_0} \cosh^3 K}{\cosh K'} \right\}$$

\item

Fixed points $K^\star$ of the RG equation satisfy

$$K^\star = \tanh^{-1} ( \tanh^3 K^\star )$$

and writing $x = \tanh K^\star$

$$x =  x^3 \Rightarrow x(1 - x^2) = 0 \Rightarrow x = -1, 0, 1$$

$\tanh^{-1} 0 = 0$ and $\tanh^{-1} (\pm1) = \pm \infty$, but $K \geq 0$ physically, so the fixed points are

$$K^\star = 0, \infty$$

Around the $K^\star = 0$ fixed point we can linearise the RG equation according to $\tanh x \sim x$ for $x \ll 1$

$$K' \sim K^3$$

so $K' < K$, and the value of $K$ decreases under successive rescalings. Above the completely disordered $K = 0$ point, under rescaling the system is driven back towards disorder;  $K^\star = 0$ is an \emph{attractive} fixed point.

To study the behaviour around the $K^\star = \infty$ fixed point, consider

$$\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{1-e^{-2x}}{1+e^{-2x}} = (1-e^{-2x})(1+e^{-2x} + \mathcal{O}(e^{4x})) = 1 + 2e^{-2x} + \mathcal{O}(e^{-4x})$$

so the RG equation around $K^\star = \infty$ is

$$1 + 2e^{-2K'} + \mathcal{O}(e^{-4K'}) = \left(1 + 2e^{-2K} + \mathcal{O}(e^{-4K}) \right)^3 = 1 + 6 e^{-2K} + \mathcal{O}(e^{-4K})$$

$$\Rightarrow e^{-2K'} = 3e^{-2K} \Rightarrow K' = -\frac{1}{2} ( \log 3 - 2K) = K - \frac{1}{2} \log 3$$

so $K' < K$ and the system is driven away from $K^\star = \infty$ under rescaling; $K^\star = \infty$ is a \emph{repulsive} fixed point.

Indeed, since there are only two (physical) fixed points, it suffices to determine the flow at one of these points to demonstrate that the overall flow must always be from $+ \infty$ towards $0$ (there are no fixed p2oints between these values).

\item

For general integer $\lambda \geq 2$ the decimation procedure gives

$$e^{K'_0} e^{K' \sigma_i \sigma_{i+\lambda}} = e^{\lambda K_0}\sum_{\sigma_{i+1} = \pm 1} \sum_{\sigma_{i+2} = \pm 1} \cdots \sum_{\sigma_{i+\lambda-1} = \pm 1} e^{K \sigma_i \sigma_{i+1}} e^{K \sigma_{i+1} \sigma_{i+2}} \cdots e^{K \sigma_{i+\lambda-1} \sigma_{i+\lambda}}$$

and again from the identity $e^{K \sigma \sigma'} = \cosh K (1 + \sigma \sigma' \tanh K)$ and the summation properties of the $\sigma$s we obtain

$$e^{K'_0} \cosh K' \left( 1 + \sigma_i \sigma_{i+\lambda} \tanh K' \right) = 2^{\lambda-1}e^{3K_0} \cosh^\lambda K \left( 1 + \sigma_i \sigma_{i+\lambda} \tanh^\lambda K \right)$$

so for the trivial field the RG equation is

$$K'_0 = \log \left\{ \frac{e^{2^{\lambda-1}  \lambda K_0} \cosh^\lambda K}{\cosh K'} \right\}$$

and for the ferromagnetic coupling

$$K' = \tanh^{-1} ( \tanh^\lambda K )$$

Solving for the fixed points, defining $x = \tanh K^\star$

$$x = x^\lambda \Rightarrow x(1-x^{\lambda-1}) = 0 \Rightarrow x = 0,1,(-1 \textrm{ for odd } \lambda)$$

$$\Rightarrow K^\star = 0, +\infty, (-\infty \textrm{ for odd } \lambda)$$

again, we require that $K \geq 0$ and so discount the extra solutions for odd $\lambda$. Around $K^\star = 0$

$$K' \sim K^\lambda$$

so the $K^\star = 0$ (completely disordered phase) fixed point is always \emph{attractive}, and likewise around $K^\star = \infty$

$$1 + 2e^{-2K'} + \mathcal{O}(e^{-4K'}) = \left(1 + 2e^{-2K} + \mathcal{O}(e^{-4K}) \right)^\lambda = 1 + 2 \lambda e^{-2K} + \mathcal{O}(e^{-4K})$$

$$\Rightarrow e^{-2K'} = \lambda e^{-2K} \Rightarrow K' = -\frac{1}{2} ( \log \lambda - 2K) = K - \frac{1}{2} \log \lambda$$

the critical point is always \emph{repulsive}, for all integer $\lambda \geq 2$.
\end{itemize}
\end{document}